18.05/17.85	MAYBE
18.05/17.85	
18.05/17.85	DP problem for innermost termination.
18.05/17.85	P =
18.05/17.85	  f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
18.05/17.85	  f10#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f5#(I15, I16, I17, I18, I19) -> f8#(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  f9#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
18.05/17.85	  f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
18.05/17.85	  f8#(I35, I36, I37, I38, I39) -> f7#(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 
18.05/17.85	  f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  f1#(I70, I71, I72, I73, I74) -> f2#(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	R = 
18.05/17.85	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
18.05/17.85	  f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
18.05/17.85	  f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
18.05/17.85	  f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46]
18.05/17.85	  f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 
18.05/17.85	  f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0]
18.05/17.85	  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67]
18.05/17.85	  f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	
18.05/17.85	The dependency graph for this problem is:
18.05/17.85	  0 -> 1
18.05/17.85	  1 -> 11
18.05/17.85	  2 -> 5, 6
18.05/17.85	  3 -> 5, 6
18.05/17.85	  4 -> 8
18.05/17.85	  5 -> 8
18.05/17.85	  6 -> 10
18.05/17.85	  7 -> 9
18.05/17.85	  8 -> 7
18.05/17.85	  9 -> 7
18.05/17.85	  10 -> 2, 3, 4
18.05/17.85	  11 -> 12
18.05/17.85	  12 -> 10
18.05/17.85	Where:
18.05/17.85	  0) f11#(x1, x2, x3, x4, x5) -> f10#(x1, x2, x3, x4, x5) 
18.05/17.85	  1) f10#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  2) f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  3) f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  4) f5#(I15, I16, I17, I18, I19) -> f8#(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  5) f9#(I20, I21, I22, I23, I24) -> f8#(I20, I21, I22, I23, I24) 
18.05/17.85	  6) f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  7) f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
18.05/17.85	  8) f8#(I35, I36, I37, I38, I39) -> f7#(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  9) f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  10) f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 
18.05/17.85	  11) f4#(I60, I61, I62, I63, I64) -> f1#(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  12) f1#(I70, I71, I72, I73, I74) -> f2#(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	
18.05/17.85	We have the following SCCs.
18.05/17.85	  { 2, 3, 6, 10 }
18.05/17.85	  { 7, 9 }
18.05/17.85	
18.05/17.85	DP problem for innermost termination.
18.05/17.85	P =
18.05/17.85	  f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
18.05/17.85	  f6#(I40, I41, I42, I43, I44) -> f7#(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	R = 
18.05/17.85	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
18.05/17.85	  f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
18.05/17.85	  f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
18.05/17.85	  f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46]
18.05/17.85	  f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 
18.05/17.85	  f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0]
18.05/17.85	  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67]
18.05/17.85	  f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	
18.05/17.85	We use the reverse value criterion with the projection function NU:
18.05/17.85	  NU[f6#(z1,z2,z3,z4,z5)] = z1 + -1 * (1 + z2)
18.05/17.85	  NU[f7#(z1,z2,z3,z4,z5)] = z1 + -1 * (1 + z2)
18.05/17.85	
18.05/17.85	This gives the following inequalities:
18.05/17.85	    ==>  I30 + -1 * (1 + I31) >= I30 + -1 * (1 + I31)
18.05/17.85	  1 + I41 <= I40  ==>  I40 + -1 * (1 + I41) > I40 + -1 * (1 + (1 + I41))  with  I40 + -1 * (1 + I41) >= 0
18.05/17.85	
18.05/17.85	We remove all the strictly oriented dependency pairs.
18.05/17.85	
18.05/17.85	DP problem for innermost termination.
18.05/17.85	P =
18.05/17.85	  f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
18.05/17.85	R = 
18.05/17.85	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
18.05/17.85	  f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
18.05/17.85	  f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
18.05/17.85	  f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46]
18.05/17.85	  f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 
18.05/17.85	  f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0]
18.05/17.85	  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67]
18.05/17.85	  f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	
18.05/17.85	The dependency graph for this problem is:
18.05/17.85	  7 -> 
18.05/17.85	Where:
18.05/17.85	  7) f7#(I30, I31, I32, I33, I34) -> f6#(I30, I31, I32, I33, I34) 
18.05/17.85	
18.05/17.85	We have the following SCCs.
18.05/17.85	  
18.05/17.85	
18.05/17.85	DP problem for innermost termination.
18.05/17.85	P =
18.05/17.85	  f5#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 
18.05/17.85	R = 
18.05/17.85	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
18.05/17.85	  f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
18.05/17.85	  f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
18.05/17.85	  f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46]
18.05/17.85	  f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 
18.05/17.85	  f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0]
18.05/17.85	  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67]
18.05/17.85	  f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	
18.05/17.85	We use the extended value criterion with the projection function NU:
18.05/17.85	  NU[f2#(x0,x1,x2,x3,x4)] = -x0 + x2 - 1
18.05/17.85	  NU[f9#(x0,x1,x2,x3,x4)] = -x0 + x2 - 2
18.05/17.85	  NU[f5#(x0,x1,x2,x3,x4)] = -x0 + x2 - 1
18.05/17.85	
18.05/17.85	This gives the following inequalities:
18.05/17.85	  1 + I5 <= I7  ==>  -I5 + I7 - 1 > -I5 + I7 - 2  with  -I5 + I7 - 1 >= 0
18.05/17.85	  1 + I12 <= I10  ==>  -I10 + I12 - 1 >= -I10 + I12 - 2
18.05/17.85	    ==>  -I25 + I27 - 2 >= -(1 + I25) + I27 - 1
18.05/17.85	    ==>  -I50 + I52 - 1 >= -I50 + I52 - 1
18.05/17.85	
18.05/17.85	We remove all the strictly oriented dependency pairs.
18.05/17.85	
18.05/17.85	DP problem for innermost termination.
18.05/17.85	P =
18.05/17.85	  f5#(I10, I11, I12, I13, I14) -> f9#(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f9#(I25, I26, I27, I28, I29) -> f2#(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f2#(I50, I51, I52, I53, I54) -> f5#(I50, I51, I52, I53, I54) 
18.05/17.85	R = 
18.05/17.85	  f11(x1, x2, x3, x4, x5) -> f10(x1, x2, x3, x4, x5) 
18.05/17.85	  f10(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4]
18.05/17.85	  f5(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, I9) [1 + I5 <= I7]
18.05/17.85	  f5(I10, I11, I12, I13, I14) -> f9(I10, I11, I12, I13, I14) [1 + I12 <= I10]
18.05/17.85	  f5(I15, I16, I17, I18, I19) -> f8(I15, I16, I17, I18, I19) [I17 <= I15 /\ I15 <= I17]
18.05/17.85	  f9(I20, I21, I22, I23, I24) -> f8(I20, I21, I22, I23, I24) 
18.05/17.85	  f9(I25, I26, I27, I28, I29) -> f2(1 + I25, I26, I27, I28, I29) 
18.05/17.85	  f7(I30, I31, I32, I33, I34) -> f6(I30, I31, I32, I33, I34) 
18.05/17.85	  f8(I35, I36, I37, I38, I39) -> f7(-1 + I35, 0, I37, I38, I39) 
18.05/17.85	  f6(I40, I41, I42, I43, I44) -> f7(I40, 1 + I41, I42, I43, I44) [1 + I41 <= I40]
18.05/17.85	  f6(I45, I46, I47, I48, I49) -> f3(I45, I46, I47, I48, I49) [I45 <= I46]
18.05/17.85	  f2(I50, I51, I52, I53, I54) -> f5(I50, I51, I52, I53, I54) 
18.05/17.85	  f4(I55, I56, I57, I58, I59) -> f3(I55, I56, I57, I58, I59) [I57 <= 0]
18.05/17.85	  f4(I60, I61, I62, I63, I64) -> f1(I60, I61, I62, I63, I64) [1 <= I62]
18.05/17.85	  f1(I65, I66, I67, I68, I69) -> f3(I65, I66, I67, I68, I69) [101 <= I67]
18.05/17.85	  f1(I70, I71, I72, I73, I74) -> f2(0, I71, I72, I73, rnd5) [rnd5 = rnd5 /\ I72 <= 100]
18.05/17.85	
18.05/20.83	EOF